There is an unexpected proof that comes from looking at the hexagon from a slightly different angle, literally. Can you find it for the win?

I think I might have it.

We can construct a stylised 3D cube out of the three diamonds in the shape of a 2D hexagon. In order to complete the cube/hexagon, we require one of each diamond orientation. For a cube of length n, there will be n^2 number of diamonds of each orientation.

I hope that makes sense. I wish I could scan and upload my drawing. Once I realised that it's a stylised cube for n=1, it took me a bit to actually draw out n=2. Once I did, though, it's easy to see that as one side of the cube grows, the others must (obviously) grow as well in order to maintain the shape of the regular hexagon (and cube).

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First, we'll look at each "ring" of hexagons. The centre hexagon contains 6 diamonds in 2 sets of 3. In order to not overlap diamonds, choosing the first one forces the orientation of the other two to be different. For each ring around the centre hexagon, the same is true. Once a single orientation is chosen, the orientation of all of the diamonds in that ring are chosen. Additionally, two adjacent edges will share an orientation before the orientation is forced to change (if you want me to prove why the orientation is forced to change, I can't come up with anything other than "you've run out of room"). This pattern continues for all rings, regardless of how many you add.

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The symmetry of course is 6-fold, so 2 is a red herring, and we can say, instead:

A regular hexagon is divided into 6p equilateral triangles.
Paring triangles that share an edge produces diamond shapes with three distinct orientations, as shown.
Prove that any tiling of the hexagon will use exactly p diamonds of each type.

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First, we'll look at each "ring" of hexagons. The centre hexagon contains 6 diamonds in 2 sets of 3. In order to not overlap diamonds, choosing the first one forces the orientation of the other two to be different. For each ring around the centre hexagon, the same is true. etc. (bn)

True, and that is a proof by construction, getting you a point.

There is an unexpected proof that comes from looking at the hexagon from a slightly different angle, literally. Can you find it for the win?